Expand $\frac{x}{4-x^2}$ as a power series around $x_0 = 0$ and compute its radius of convergence.
$\frac{x}{4-x^2} = x \frac1{4-x^2} = \frac{x}4 \frac{1}{1-\frac{x^2}{4}} = \frac{x}4\sum_{n=0}^\infty (\frac{x^2}{4})^n \text{(assuming that} \frac{x^2}{4} <1) = \sum_{n=0}^\infty \frac1{4^{n+1}} x^{2n+1}$.
I am stuck here. How can I proceed from here to get the power series : $\sum_{n=0}^\infty a_n (x – 0)^n$?
Best Answer
Use
$$\frac1{1-t} = 1+ t+t^2+t^3+\>...$$
to expand
$$\frac{x}{4-x^2}=\frac x4\cdot\frac1{1-(\frac x2)^2} =\frac x4\cdot \left( 1 + (\frac x2)^2 + (\frac x2)^4+(\frac x2)^6+\>... \right)$$ $$=\frac x{2^2} + \frac {x^3}{2^4} + \frac {x^5}{2^6}+\frac {x^7}{2^8}+\>... = \sum_{k=0}^\infty \frac{[1-(-1)^k]}{2^{k+2}}x^k $$
Its radius of convergence is $(\frac x2)^2 < 1$, or $|x|<2$.